# Does the following series converge uniformly in $[1,\infty)$? $\sum_{n=1}^\infty \frac{x^n}{(1+x)(1+x^2)\cdots (1+x^n)}$

Does the following series converge uniformly in $[1,\infty)$? $$\sum_{n=1}^\infty \frac{x^n}{(1+x)(1+x^2)\cdots (1+x^n)}$$

I thought I might be able to prove it using Weierstrass criterion (for uniform convergence), but all I go to was a lower bound of $1^n$ which does not converge.

## 2 Answers

We can suppose $n\geq 2$. Since $x^n/(1+x^n)\leq 1$ and $1+x^m \geq 2$ $\forall m \geq 0$, we have

$$\frac{x^n}{(1+x)(1+x^2)\cdots (1+x^n)} \leq \frac{1}{(1+x)\cdots(1+x^{n-1})} \leq \frac{1}{2^{n-1}}$$

and the result is clear by using Weierstrass M-Test.

• Pretty simple actually. Nice job Thanks :) – Jason Jul 1 '18 at 13:01

Since Rafael Gonzalez Lopez already provided a proof for uniform convergence, I would like to note that the series converges uniformly to $1$ on $[1,\infty)$. To prove this, observe that $$\frac{x^n}{\prod_{j=1}^n\,\left(1+x^j\right)}=\frac{1}{\prod_{j=1}^{n-1}\,\left(1+x^j\right)}-\frac{1}{\prod_{j=1}^n\,\left(1+x^j\right)}\text{ for each }x>0\,.$$ That is, for each $m=1,2,3,\ldots$, we have $$\sum_{n=1}^m\,\frac{x^n}{\prod_{j=1}^n\,\left(1+x^j\right)}=1-\frac{1}{\prod_{j=1}^m\,\left(1+x^j\right)}\,.$$ The claim follows immediately.